2022 AMC 10A Problem 16

Below is the professionally curated solution for Problem 16 of the 2022 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2022 AMC 10A solutions, or check the answer key.

All of the real AMC 8, AMC 10, AMC 12, and AIME problems in our complete solution collection are used with official legal permission of the Mathematical Association of America (MAA).

Concepts:Vieta’s Formulaspolynomialvolume

Difficulty rating: 1420

16.

The roots of the polynomial 10x339x2+29x610x^3 - 39x^2 + 29x - 6 are the height, length, and width of a rectangular box (right rectangular prism). A new rectangular box is formed by lengthening each edge of the original box by 22 units. What is the volume of the new box?

245\dfrac{24}{5}

425\dfrac{42}{5}

815\dfrac{81}{5}

3030

4848

Solution:

Let h,l,h, l, and ww be the dimensions of the old box. Then the volume of the new box is (h+2)(l+2)(w+2). (h + 2)(l + 2)(w + 2). Expanding, we get hlw+2(hl+hw+lw) hlw + 2(hl + hw + lw) +4(l+h+w)+8.+ 4(l + h + w) + 8. We can use Vieta's formulas to find the terms in this expression. We get that hlw=DA=35, hlw = -\dfrac{D}{A} = \dfrac{3}{5}, hl+hw+lw=CA=2910, hl + hw + lw = \dfrac{C}{A} = \dfrac{29}{10}, and l+h+w=BA=3910. l + h + w = -\dfrac{B}{A} = \dfrac{39}{10}.

Plugging these values into the expression, we get 35+22910+43910+8=30. \dfrac{3}{5} + 2 \cdot \dfrac{29}{10} + 4 \cdot \dfrac{39}{10} + 8 = 30.

Thus, D is the correct answer.

Problem 16 in Other Years